How is the integral of int x / ( sinh(x) + cosh(x) ) dxxsinh(x)+cosh(x)dx?

2 Answers
May 16, 2018

-xe^(-x)-e^(-x)+Cxexex+C

Explanation:

int x/(sinhx+coshx)*dxxsinhx+coshxdx

=int x/(e^x)*dxxexdx

=int xe^(-x)*dxxexdx

=x*(-e^(-x))-int (-e^(-x))*dxx(ex)(ex)dx

=-xe^(-x)+int e^(-x)*dxxex+exdx

=-xe^(-x)-e^(-x)+Cxexex+C

May 16, 2018

Given: int x / ( sinh(x) + cosh(x) ) dxxsinh(x)+cosh(x)dx

Multiply the integrand by 1 in the form of (sinh(x)-cosh(x))/(sinh(x)-cosh(x))sinh(x)cosh(x)sinh(x)cosh(x):

int x / ( sinh(x) + cosh(x) ) dx = int x / ( sinh(x) + cosh(x) )(sinh(x)-cosh(x))/(sinh(x)-cosh(x)) dxxsinh(x)+cosh(x)dx=xsinh(x)+cosh(x)sinh(x)cosh(x)sinh(x)cosh(x)dx

The denominator is multiplied using the difference of two squares pattern and the numerator is multiplied using the distributive property:

int x / ( sinh(x) + cosh(x) ) dx = int (xsinh(x)-xcosh(x)) / ( sinh^2(x) - cosh^2(x)) dxxsinh(x)+cosh(x)dx=xsinh(x)xcosh(x)sinh2(x)cosh2(x)dx

The identity cosh^2(x)- sinh^2(x)=1cosh2(x)sinh2(x)=1 tells us that the denominator is -1:

int x / ( sinh(x) + cosh(x) ) dx = int (xsinh(x)-xcosh(x)) / -1 dxxsinh(x)+cosh(x)dx=xsinh(x)xcosh(x)1dx

Eliminate the denominator by changing signs in the numerator:

int x / ( sinh(x) + cosh(x) ) dx = int xcosh(x)-xsinh(x) dxxsinh(x)+cosh(x)dx=xcosh(x)xsinh(x)dx

Separate into two integrals:

int x / ( sinh(x) + cosh(x) ) dx = int xcosh(x) dx- int xsinh(x) dxxsinh(x)+cosh(x)dx=xcosh(x)dxxsinh(x)dx

Both integrals are trivial integrations by parts:

int x / ( sinh(x) + cosh(x) ) dx = (xsinh(x) - cosh(x)) - (xcosh(x)-sinh(x))+Cxsinh(x)+cosh(x)dx=(xsinh(x)cosh(x))(xcosh(x)sinh(x))+C

Regroup:

int x / ( sinh(x) + cosh(x) ) dx = xsinh(x)- xcosh(x)+ sinh(x)- cosh(x) +Cxsinh(x)+cosh(x)dx=xsinh(x)xcosh(x)+sinh(x)cosh(x)+C